The angle of elevation of the top of a cliff 15 meters high from a landmark is 60°. How far is the landmark from the foot of the cliff? Leave your answer in surd form
\(15\sqrt{3}m\)
\(15\sqrt{2}m\)
\(10\sqrt{3}m\)
\(5\sqrt{3}m\)
Correct answer is A
Tan 60º = \(\frac{15}{x}\)
cross multiply:
x = \(\frac{15}{\sqrt3}\) = 5\(\sqrt3\)m